Multiple Choice
A 30.00-mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 21.4 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10⁻⁴
289
views
18. Aqueous Equilibrium
Titrations: Weak Acid-Strong Base
Multiple Choice
A 35.0-mL sample of 0.150 M acetic acid (CH3COOH, Ka=1.8×10⁻⁵) is titrated with 0.150 M NaOH solution. Calculate the pH after 35.0 mL of the base have been added.
A
4.76
B
8.72
C
7.00
D
3.75
Verified step by step guidance1
Determine the initial moles of acetic acid (CH3COOH) using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). Here, the concentration is 0.150 M and the volume is 35.0 mL (convert to liters by dividing by 1000).
Calculate the moles of NaOH added using the same formula: \( \text{moles} = \text{concentration} \times \text{volume} \). The concentration of NaOH is 0.150 M and the volume is 35.0 mL (convert to liters).
Since the moles of acetic acid and NaOH are equal, they will completely neutralize each other, forming water and the acetate ion (CH3COO⁻). This is a stoichiometric point where the number of moles of acid equals the number of moles of base.
At the equivalence point in a titration of a weak acid with a strong base, the solution consists of the conjugate base (CH3COO⁻) of the weak acid. The pH is determined by the hydrolysis of the acetate ion: \( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \).
Calculate the pH by determining the concentration of OH⁻ produced from the hydrolysis of the acetate ion and then using the relation \( \text{pH} = 14 - \text{pOH} \). Since the solution is neutral at the equivalence point for a weak acid-strong base titration, the pH is approximately 7.00.
Related Videos
Related Practice
